• Coplanarity constraint
• Intersection of two corresponding rays
• The rays lie in a 3D plane

Epipolar Geometry

Epipolar Geometry

Epipolar elements:

We can also write the epipoles as:

$$e' = (O' O'') \cap \mathcal{E}', \quad e'' = (O' O'') \cap \mathcal{E}''$$

And the epipolar lines as:

$$\mathcal{L}'(X) = \mathcal{E} \cap \mathcal{E}', \quad \mathcal{L}''(X) = \mathcal{E} \cap \mathcal{E}''$$

In the Epipolar Plane

Assuming a distortion free lens:

In the Epipolar Plane

• The projection centres $O'$ and $O''$.
• The observed point $X$.
• The epipolar lines, $\mathcal{L}'(X)$ and $\mathcal{L}''(X)$.
• The epipoles, $e'$ and $e''$.
• The image points $x'$ and $x''$.

All lie in the epipolar plane $\mathcal{E}$.

Predicting Point Correspondence

Task: Predict the location of $x''$ given $x'$.

• For the epipolar plane $\mathcal{E} = (O'O''X)$
• The intersection of $\mathcal{E}$ and the second image plane $\mathcal{E}''$ yields the epipolar line $\mathcal{L}''(X)$
• The corresponding point $x''$ lies on that epipolar line $\mathcal{L}''(X)$.
• Search space is reduced from 2D to 1D.

Computing the Elements of Epipolar Geometry

• We described the important elements geometrically.
• We will compute them using the projection matrices and the fundamental matrix.

Epipolar Axis

The direction of the epipolar axis can be derived directly from the projection centres.

$$b = X_{O'} - X_{O''}$$

The vector $b$ is homogeneous, we know only the direction, not the length.

Epipolar Lines

Image points lie on the epipolar lines.

$$x' \in \mathcal{L}' , \quad x'' \in \mathcal{L}''$$

Epipolar Lines

For $x'$:

$$x^{'T} \mathcal{L}' = 0$$

Epipolar Lines

For $x'$:

$$x^{'T} \mathcal{L}' = 0$$

We can exploit the coplanarity constraint for $x'$ and $x''$:

$$x^{'T} \underbrace{Fx''}_{\mathcal{L}'} = 0$$

Epipolar Lines

We can exploit the coplanarity constraint for $x'$ and $x''$:

$$x^{'T} \underbrace{Fx''}_{\mathcal{L}'} = 0$$

$$\mathcal{L}' = Fx''$$

Epipolar Lines

The same for $x''$:

$$\mathcal{L}^{''T} x'' = 0$$

We can exploit the same constraint $x^{'T}Fx''= 0$ and obtain:

$$\mathcal{L}^{''T} = x^{'T}F$$

Epipolar Lines

We can exploit the same constraint $x^{'T}Fx''= 0$ and obtain:

$$\mathcal{L}^{''T} = x^{'T}F$$

$$\mathcal{L}^{''} = F^{T}x^{'}$$

Epipolar Lines

Image points lie on the epipolar lines, $x' \in \mathcal{L}'$ and $x'' \in \mathcal{L}''$.

• we can exploit the coplanarity constraint $x^{'T}Fx''= 0$.
• which is valid if:

$$\mathcal{L}' = Fx'' \quad \mathcal{L}'' = F^{T}x'$$

Epipoles

The epipoles are the projection of the camera origin onto the other image.

• Both can be computed using the projection matrices.

Epipoles

The epipoles are the projection of the camera origin onto the other image.

• Both can be computed using the projection matrices.

$$e' = P'X_{O''} \quad e'' = P''X_{O'}$$

Epipoles

The epipole is the intersection of all the epipolar lines in an image.

$$\forall \mathcal{L}' : e^{'T} \mathcal{L}' = 0 \quad \forall \mathcal{L}'' : e^{''T} \mathcal{L}'' = 0$$